mechanics of solids (me f211) chaudhari/mechanics of... · mechanics of solids chapter-2...

BITS Pilani K K Birla Goa Campus VIKAS CHAUDHARI

MECHANICS OF SOLIDS

(ME F211)

Vikas Chaudhari BITS Pilani, K K Birla Goa Campus

Mechanics of Solids

Chapter-2

Introduction to Mechanics of deformable bodies



Analysis of deformable bodies

Identification of a system

Simplification of this system

To develop model which can be analyzed

Steps to analyze system

Study of forces and equilibrium requirements

Study of deformation and conditions of geometric fit

Application of force- deformation relations



Example:

A machine part carrying a load F terminates

in a piston which fits into a cavity, as shown

in Fig. Within the cavity are two linear

springs arranged coaxial with each other.

We use the symbols kA and kB to denote the

spring constants of the two springs in the

cavity. When the springs are unloaded, each

has the same length L. We wish to know

how much of the load F is carried by the

spring with constant kA



Solution

Steps to analyze the system

1. Selection of model of the actual system

2. Assumptions

3. Study of forces and equilibrium requirements

4. Study of deformation and conditions of geometric fit

5. Application of force- deformation relations to predict behavior of

the system



1. Selection of model

a. Identify the elements of proposed model of the system. In

given problem the system has two elements one Piston and

two springs

2. Assumptions

a. Assumed that the springs have been made with flat ends

such that the compressing force which is distributed around

the periphery of the spring, can be considered to act along

the spring axis

b. Assumed that gravity effects can be ignored

c. Springs fit in the cavity perfectly without buckling



3. Study of forces and equilibrium requirements

a. Isolate the elements from its surrounding i.e. Draw free body

diagram

b. Apply equations of equilibrium i.e. ∑F = 0 and ∑M = 0

Fy = 0 = FA + FB F --- 1

Free body Diagram




a. Find what are the requirements for geometric compatibility

A = B = --- 2

5. Application of force- deformation relations

a. The force in each spring is linearly proportional to the deflection

of the spring, and the constant of proportionality is the spring

constant

FA = kA A --- 3

FB = kB B --- 4



By combining equation 1-4, we obtained desire result

and A A B B

A B A B

F k F k

F k k F k k

A B

A A B B

F FF

k k k k

Total deflection of piston



Example:

Let us consider a very light and stiff

wood plank of length 2L. Two similar

springs (spring constant k) are

attached to it. The springs are of

length h when the plank is resting on

them. Let a man stands on middle of

plank and begins to walk slowly

towards one end. Find how far the

man can walk before one end of the

plank touches the ground



Solution:

1. Force equilibrium : Free body diagram

Reaction at point E will be zero since the plank is just touching the ground

∑Fy = 0 FC + FD − W = 0 --- 1

∑M@C = 0 2 a FD − (a + b) W = 0 --- 2

c




When the plank remains straight, from the similar triangles

--- 3C

D

h L a

h L a

The deflections of the springs are

--- 4C C D Dh h and h h

d



3. Relations between forces and deflections

Since both springs are linear and have same spring constant,

2 21 --- 6

a khb

L W

Equation 1 to 5 will give seven independent relations for seven unknowns, FC, FD, hC, hD, C, D, and b.

Solving these equations, we find that the value of b is given by

--- 5C C D DF k and F k



Spring deflections

1 and 1 --- 72 2

C D

W b W b

k a k a

Highlights

It can be seen that D is always positive in the sense defined in Fig d.

C is positive as long as b < a. When b = a, the man is directly over the

spring D, and, as would be expected, all the load is taken by the spring

D, and the deflection and force in the spring C are zero.

When b > a, then C is negative (i.e., the spring extends). In Fig. c we

assumed that b > a and that the spring C is compressed.



UNIAXIAL LOADING & DEFORMATION

The basic type of deformation which is

considered in most of the problems in

this chapter is shown in Figure (a).

Consider the deformation of three rods

of same material, but different lengths

and cross-sectional areas as shown.

Assume that for each bar the load is

gradually increased from zero, and at

several values of the load a elongation

() is measured.




If the maximum elongation is very small (less than 0.1 percent of

the original length), then for most materials the results of the three

tests will be represented by a plot like Figure (b) or like Figure (c).




If the experimental data are re-plotted with load over area (P/A)

as ordinate and elongation over original length (/L) as abscissa,

the test results for the three bars can be represented by a single

curve, as shown in Figure (d) or Figure (e).




If the uniaxial load-elongation relation of the material is linear as

shown in Figure (d), then the slope of the curve is called the

modulus of elasticity and is denoted by the symbol E

PPLAE

AL

The dimension of E is N/m2 or N/mm2.

Deflection can be given in terms of E

PL

AE



Statically determinate situations

These are the situations in which the forces can be obtained without

reference to the geometry of deformation

Example

Figure shows a triangular frame supporting a

load of 20kN. Our aim is to estimate the

displacement at the point D due to the 20kN

load carried by the chain hoist.



Solution

1. Force equilibrium:

The forces in the members were determined in previous chapter

FBD = 28.3kN Tensile FCD = 20kN Compressive



2. Force deformation relation

3 3

3

3 3

3

28.3 10 3 2 101.19 --- extension

491 205 10

20 10 3 100.0915 --- compression

3200 205 10

BD

BD

CD

CD

PLmm

AE

PLmm

AE

3. Geometric compatibility

Bars BD and CD move in such a way that, after deformation, they remain

straight and also remain fastened together at D as shown in Figure (c)

Since the deformations of the bars are very small fractions of the lengths,

we can replace the arcs by the tangents to the arcs at D1 and D2 and

obtain the intersection D4 as an approximation to the location of D3 Fig (d).



H = CD = 0.0915mm

V = D2F + FD4 = DG + FG = 2 BD + CD = 1.77mm



Example

The stiff horizontal beam AD in Figure (a) is supported by two soft

copper rods AC and BD of the same cross-sectional area but of

different lengths. The load-deformation diagram for the copper is

shown In Figure (b). A vertical load of 150kN is to be suspended from

a roller which rides on the horizontal beam. We do not want the

roller to move after the load is put on, so we wish to find out where

to locate the roller so that the beam will still be horizontal in the

deflected position.



1. Force equilibrium

Fy = 0 = FA + FB 150 --- 1

M@A = 0 = FB 1 150 c --- 1(A)


A = B --- 2

from the lengths of the bars

2 2 --- 31.3 2.6

A B B B B

A A BL L L



3. Force-deformation relation

In given problem, force and deformation are

related with Figure (b).

e.g. if we choose B/LB in diagram equal to

0.001 then corresponding value of FB/AB

would be 74 MN/m2. --- 4

Equations 1, 3 and 4 represent our analysis of

the physics of the problem. We must combine

1, 3, and 4 mathematically to find the correct

location of the roller.



Dividing equation 1 by AA and substituting AA = AB = 1300mm2

2

150 115 --- 5A B A B

A A A A B

F F F F MN

A A A A A m

Procedure to proceed further

1. Select an arbitrary value of B/LB.

2. Using equation 3, obtain A/LA.

3. Obtain FB/AB and FA/AA from Figure (b) as explained in equation 4.

4. Check whether these values satisfy equation 5.

5. If equation 5 is not satisfied, make a new guess for B/LB and obtain

new values for A/LA, FA/AA and FB/AB.



6. Proceeding in this way, we find the points a and b in Figure (b). From

these points,

2

2

0.0005 41 / 53.3

0.001 74 / 96.2

B B

B

B B

A A

A

A A

FMN m F kN

L A

FMN m F kN

L A

A = B = 1.3mm

Substituting this value for FB in the equation 1(A), we obtain the

required location of the roller

c = 0.355m



Analysis of thin ring

Example

A thin ring of internal radius r,

thickness t, and width b is subjected

to a uniform pressure p over the

entire internal surface, as shown in

figure below. We would like to

determine the forces in the ring and

deformation of the ring due to the

internal pressure.




FR must be zero because Newton’s third law and symmetry argument, both together are not satisfied at a time.

Let’s consider upper half of the ring

[ ( )] --- (a)pF p b r

sin [ ( )]sin -----(b)y pF F p b r

0

sin( ) 2 0 ---- (c)y TF pbr d F

---- (d)TF prb

(2 ) 2 0 ---- (e)y TF p rb F



2. Force-deformation relation

2[2 ( )] 22 (1 ) ----- (f)( ) 2

T

T

tF r pr t

bt E tE r

--- (g)2

T

R

2

(1 ) ---- (h)2

R

pr t

tE r

2

----- (i)Rpr

tE




Problem

A composite hoop consists of a brass hoop of 300-mm internal radius and 3-mm thickness, and a steel hoop of 303-mm internal radius and 6-mm radial thickness. Both hoops are 6-mm thick normal to the hoop. If radial pressure of 1.4 MN/m2 is put in the brass hoop, estimate the tangential forces in the brass and steel hoops.



Example In a test on an engine, a braking force is supplied through a lever arm EF to a steel brake CBAD which is in contact with half the circumference of a 600mm diameter flywheel. The brake band is 1.6mm thick and 50mm wide and is lined with a relatively soft material which has a kinetic coefficient of friction of f = 0.4 with respect to the rotating flywheel. The operator wishes to predict how much elongation there will be in the section AB of the brake band when the braking force is such that there is a tension of 40kN in the section BC of the band.

Analysis of flexible drives




sin ( )sin 0 --- 12 2

rF N T T T

( )cos cos 0 --- 32 2

F T T T f N

( ) 0 --- 22 2

N T T T N T

( ) 0 --- 4T T T f N T f N

--- 5T dT

T fT f fdT T

From 2 & 4



At = 0; T = TAD and = π; T = TBC

0

--- 6BC

AD

T

T

dTf d

T

where e is the base of natural logarithms. TAD can be calculated by substituting TBC = 40kN and f = 0.4 in equation 7.

Suppose, if we assume T1 is tension in tight side; T2 is tension in slack side and is angle of contact in radians, then equation 7 can be written in the following form

--- 7fBC

AD

Te

T

1

2

--- 8fT

eT



2. Force-deformation relation Consider the deflection of an small element of length R and cross-section area A

--- 9

TR TR dd

AE AE

0 0

--- 10AB

TR dd

AE

3. Geometric compatibility Total elongation of the brake band from A to B, AB, is the sum of the tangential elongations of small elements of length R .

0

( 1) --- 11f fAD ADABT R T R

e d eAE AEf



Problem

A hawser from a ship is wrapped four times around a rotating capstan as shown in the figure. The dockworker pulls with a force of 200 N. What is the maximum force the man can exert on the boat if the coefficient of friction between the capstan and hawser is 0.3?



Problem

A brake is designed as shown. A 25 x 1.5 mm steel band restrains the wheel from turning when a 225 N-m torque is applied. The friction coefficient is 0.4. Find the tensions T1 and T2 that just keep the wheel from rotating.



Statically indeterminate situations

These are the situations in which deformations must be considered

to determine forces.

Example

Figure shows the pendulum of a clock which has

a12N weight suspended by three rods of 760mm

length. Two of the rods are made of brass and the

third of steel. We wish to know how much of the

12N suspended weight is carried by each rod.

Take ES = 200GPa and EB = 100 Gpa.



Solution

1. Force equilibrium:

12 2 0 --- 1y S BF F F

2. Geometric compatibility:

--- 2S B

3. Force-deformation relation:

and --- 3S S B BS BS S B B

F L F L

A E A E

From equations 1 to 3

2.55 and 4.72S BF N F N



Problem

A square reinforce-concrete pier

0.3 0.3m in cross section and

1.2m high is loaded as shown in

the figure. The concrete is

strengthened by the addition of

eight vertical 25 25mm square

steel reinforcing bars placed

symmetrically about the vertical

axis of the pier. Find the stress

(force per unit area) in the steel

and concrete and the deflection.

Take EC = 17GPa and ES = 200GPa.



ELASTIC ENERGY; CASTIGLIANO'S THEOREM

Conservative systems:

When work is done by an external force on certain systems, their

internal geometric states are altered in such a way that they have

the potential to give back equal amounts of work whenever they are

returned to their original configurations.

e.g. Elastic spring



Strain Energy:

Consider the elastic, but not necessarily linear, spring in Fig (a). Let the spring undergo a gradual elongation process during which the external force F remains in equilibrium with the internal tension. The potential or strain energy U associated with an elongation is defined to be the work done by F in this process

0

U Fd



Complementary Energy:

When the point of application of a variable force F undergoes a displacement , the complementary work is done on the system.

When complementary work is done on certain systems, their internal force states are altered in such a way that they are capable of giving up equal amounts of complementary work when they are returned to their original force states. Under these circumstances the complementary work done on such a system is said to be stored as complementary energy.

*

0

U dF



Castigliano’s theorem:

It states that if the total complementary energy U* of a loaded

elastic system is expressed in terms of the loads, the in-line

deflection at any particular loading point is obtained by

differentiating U* with respect to the load at that point.

*U

P

For nonlinear systems, U U*

For linear systems, U = U*

Strain Energy of linear spring is 2

21 1 1

2 2 2

FU F k

k



This chapter, we will consider application of castigliano’s

theorem only for linear elastic systems.

For linear system, complementary energy is equal to the strain

energy.

Let us consider a system, consists of N elastic elements. External

force Pi acting at point Ai. We wish to know the deflection of any

point Ai.

Total strain energy of the system is where Fi is the internal force induced in elastic member and it should be expressed in terms of Pi

2

1

1

2

ni

i i

FU

k



Deformation (i), at point Ai in the direction of Pi is given by

i

i

U

P

Important

If, we may wish to know deflection (may be in horizontal or

vertical or in any other direction) at a point where external force

is zero .

In such case a fictitious force Q is to be considered at that point.

Express internal forces in terms of Q.

Deflection at that point in the direction of Q is given by U/Q

and setting Q = 0.



Example

Consider the system of two springs shown in Figure and determine

the deflections using Castigliano's theorem.



Solution

From equilibrium requirements, the spring forces (F1 and F2) are

expressed as

F1 = P1 + P2 & F2 = P2 The total elastic energy is given by

2 2

1 2 2

1 2

1 2

( )

2 2

P P PU U U

k k

1 2 1 2 21 2

1 1 2 1 2

and P P P P PU U

P k P k k

The deflections (1 and 2) are



Example

Figure shows a triangular

frame supporting a load of

20kN. Using Castigliano's

theorem, determine vertical

and horizontal deflection of

point D.



Solution

Vertical force say P is acting at point D

(P = 20kN).

Horizontal force is zero at point D.

Lets consider a fictitious force Q acting at

point D in horizontal direction as shown in

figure.

Express FBD and FCD in terms of P and Q.



From FBD of joint D,

0 sin 45 2Y BD BDF F P F P

0 cos 45 X CD BD CDF Q F F F Q P

Total strain energy, U of the system is

222 22

2 2 2 2

CDBD

BD CD

BD CD BD CD

Q PFF PU U U

k k k k

where kBD and kCD are stiffness of bar BD and CD

33

3

33

3

491 205 1023.72 10 /

3 2 10

3200 205 10218.67 10 /

3 10

BD

BD

BD

CD

CD

CD

A Ek N mm

L

A Ek N mm

L



Vertical deflection of point D,

Setting P = 20kN and Q = 0 24

2 2VD

BD CD

P QU P

P k k

333 3

2 20 10 04 20 101.77

2 23.73 10 2 218.66 10VD mm

horizontal deflection of point D,

33

2 0 20 1020 0.091

2 2 218.66 10HD

CD

Q PUmm

Q k

-ve sign indicates that the horizontal deflection of point D is in

opposite to the direction of Q i.e. rod CD gets compressed.



Problem

A very stiff horizontal member is supported

by two vertical steel rods of different cross

section area and length. If a vertical load of

120kN is applied to the horizontal beam at

point B, estimate the vertical deflection of

the point B.


References

1. Introduction to Mechanics of Solids by S. H. Crandall et al

(In SI units), McGraw-Hill


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